$ A = \left[\begin{array}{rrr}3 & 5 & 4 \\ -1 & 4 & 5\end{array}\right]$ $ E = \left[\begin{array}{rr}5 & 4 \\ -1 & 0 \\ 5 & 5\end{array}\right]$ What is $ A E$ ?
Solution: Because $ A$ has dimensions $(2\times3)$ and $ E$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ A E = \left[\begin{array}{rrr}{3} & {5} & {4} \\ {-1} & {4} & {5}\end{array}\right] \left[\begin{array}{rr}{5} & \color{#DF0030}{4} \\ {-1} & \color{#DF0030}{0} \\ {5} & \color{#DF0030}{5}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ A$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ A$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ A$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{3}\cdot{5}+{5}\cdot{-1}+{4}\cdot{5} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ A$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{5}+{5}\cdot{-1}+{4}\cdot{5} & ? \\ {-1}\cdot{5}+{4}\cdot{-1}+{5}\cdot{5} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ A$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{3}\cdot{5}+{5}\cdot{-1}+{4}\cdot{5} & {3}\cdot\color{#DF0030}{4}+{5}\cdot\color{#DF0030}{0}+{4}\cdot\color{#DF0030}{5} \\ {-1}\cdot{5}+{4}\cdot{-1}+{5}\cdot{5} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{3}\cdot{5}+{5}\cdot{-1}+{4}\cdot{5} & {3}\cdot\color{#DF0030}{4}+{5}\cdot\color{#DF0030}{0}+{4}\cdot\color{#DF0030}{5} \\ {-1}\cdot{5}+{4}\cdot{-1}+{5}\cdot{5} & {-1}\cdot\color{#DF0030}{4}+{4}\cdot\color{#DF0030}{0}+{5}\cdot\color{#DF0030}{5}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}30 & 32 \\ 16 & 21\end{array}\right] $